# Homework Help: Proving that the intersection of two sets is a subset of Q?

1. Mar 8, 2017

### Eclair_de_XII

1. The problem statement, all variables and given/known data
"Let $U={p+r\sqrt{2}:p,r∈ℚ}$ and let $V={a+b\sqrt{7}:a,b∈ℚ}$. Show that $ℚ⊆U∩V$. Then show that $U∩V⊆ℚ$ and conclude that $U∩V=ℚ$."

2. Relevant equations

3. The attempt at a solution
I only knew how to prove the first part. That is:

(1)
"Suppose that $ℚ⊄U∩V$. This implies that $∃x∈ℚ$ such that $x∉U$ and $x∉V$. But since $ℚ⊂U$ and $ℚ⊂V$, it follows that there does not exist an $x∈ℚ⊂U$ such that $x∉U$. Similarly, there exist no $x∈ℚ⊂V$ such that $x∉V$."

I'm also ambivalent about saying that $ℚ⊂U$ when I haven't proven yet that $∃q$ such that $p+q\sqrt{2}∉ℚ$. I'm thinking about just changing it to $ℚ⊆U$.

(2)
${p+r\sqrt{2}:p,r∈ℚ}∩{a+b\sqrt{7}:a,b∈ℚ}$

And here is where I got stuck; proving that if $r≠0$, then $p+r\sqrt{2}∉V$. Similarly, I was unable to prove that $a+b\sqrt{7}∉U$. Basically, I'm trying to prove that every element of an intersection whose elements I am unable to properly identify. I'm thinking of just doing it one piece at a time:

"Suppose $p+r\sqrt{2}∈V$. Then we can say that $p+r\sqrt{2}=a+b\sqrt{7}$. With algebra, I get: $\sqrt{2}=\frac{1}{r}(s+b\sqrt{7})$ where $s=a-p$. Squaring both sides gives me $2=\frac{1}{r}(s^2+2bs\sqrt{7}+7b^2)$. However, contradicts the fact that $2$ is a prime number, since it can only be written as the product of two integers: itself and $1$."

That's only proving that any element of $U$ with $r≠0$ is an element of $V$, and I don't even think I'd be done, since I cannot rule out the possibility that $s+b\sqrt{7}=1$ and $\frac{1}{r}=2$; or the possibility that $\frac{1}{r}=1$ and $s+b\sqrt{7}=\sqrt{2}$.

I was given a hint that I needed to use the fact that $\sqrt{14}∉ℚ$ for the second part...

2. Mar 8, 2017

### Staff: Mentor

Think about what the elements of $U \cap V$ can be. It's also helpful to think about things from a vector space perspective. If you have a nonzero element q of $\mathbb{Q}$ (a rational number), any other rational number is some (rational) multiple of q. In linear algebra terms, q spans $\mathbb{Q}$. Set U can be thought of as all linear combinations of elements from $\mathbb{Q}$ and $\sqrt 2$. With set U, you get all possible rational numbers, plus something else -- all rational multiples of $\sqrt 2$. Set V is something like this, with all possible rational numbers, plus all rational multiples of $\sqrt 7$.

So I'm really thinking of U as being spanned by two "vectors" -- 1 and $\sqrt 2$, and V being spanned by a different set of vectors -- 1 and $\sqrt 7$. What do the two sets have in common?

3. Mar 8, 2017

### vela

Staff Emeritus
That should be $x∉U$ OR $x∉V$.

In any case, your proof seems unnecessarily complicated. If $x \in \mathbb{Q}$, what can you say about $x + 0\sqrt{2}$ and $x+0\sqrt{7}$?

4. Mar 9, 2017

### Eclair_de_XII

They have $1$ as that common element. So how would I express the fact that the sets are in different planes, except at the $ℚ$ axis (so to speak) where they intersect?

They're both elements of $U$ and $V$.

5. Mar 9, 2017

### vela

Staff Emeritus

If $x \in U\cap V$, you have $x = p+r\sqrt{2} = a+b\sqrt{7}$. Rearrange terms to get $p-a = b\sqrt{7}-r\sqrt{2}$ and then square both sides. See where that takes you.

6. Mar 9, 2017

### Eclair_de_XII

Let's see...

$p^2-2pa+a^2=7b^2-2br\sqrt{14}+2r^2$
$-\frac{p^2-2pa+a^2-7b^2-2r^2}{2br}=\sqrt{14}$

So I get the contradiction that $\sqrt{14}$ is supposed to be irrational? Or is it to early to say that?

Last edited: Mar 9, 2017
7. Mar 9, 2017

### Staff: Mentor

???
$\sqrt{14}$ is irrational. The left side, $p^2 - 2pa + a^2$ is rational, as are $7b^2$ and $2r^2$. Can you have a rational number on one side of an equation, and an irrational one on the other side?

8. Mar 9, 2017

### Eclair_de_XII

No; and that's where I would get the contradiction from. Okay, I think I got my answer. Thanks.

Last edited: Mar 9, 2017